Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 12}{x + 7} = \dfrac{-10x - 33}{x + 7}$
Multiply both sides by $x + 7$ $ \dfrac{x^2 - 12}{x + 7} (x + 7) = \dfrac{-10x - 33}{x + 7} (x + 7)$ $ x^2 - 12 = -10x - 33$ Subtract $-10x - 33$ from both sides: $ x^2 - 12 - (-10x - 33) = -10x - 33 - (-10x - 33)$ $ x^2 - 12 + 10x + 33 = 0$ $ x^2 + 21 + 10x = 0$ Factor the expression: $ (x + 3)(x + 7) = 0$ Therefore $x = -3$ or $x = -7$ At $x = -7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -7$, it is an extraneous solution.